Question 1189369
**a. Find an expression for q in terms of p**

The sum of all probabilities in a probability distribution must equal 1. Therefore,

q + 4p² + p + 0.7 - 4p² = 1

q + p + 0.7 = 1

q = 1 - 0.7 - p

**q = 0.3 - p**

**b. Find the value of p which gives the largest value of E(X)**

The expected value of X, E(X), is given by:

E(X) = 0*q + 1*(4p²) + 2*p + 3*(0.7 - 4p²)

E(X) = 4p² + 2p + 2.1 - 12p²

E(X) = -8p² + 2p + 2.1

To find the value of p that maximizes E(X), we can take the derivative of E(X) with respect to p and set it to zero.

d(E(X))/dp = -16p + 2

Setting the derivative to zero:

-16p + 2 = 0

16p = 2

p = 2/16

**p = 0.125**

To confirm that this is a maximum, we can take the second derivative:

d²(E(X))/dp² = -16

Since the second derivative is negative, this confirms that p = 0.125 gives a maximum value for E(X).

**c. Hence find the largest value of E(X)**

Substitute p = 0.125 into the expression for E(X):

E(X) = -8(0.125)² + 2(0.125) + 2.1

E(X) = -8(0.015625) + 0.25 + 2.1

E(X) = -0.125 + 0.25 + 2.1

**E(X) = 2.225**

Therefore, the largest value of E(X) is 2.225.